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-BARSIC- [3]
2 years ago
11

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves

with speed v = 16.24 m/s. The loop-the-loop has a radius of R = 10.5 m.
What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)

2) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3) What is the magnitude of the normal force on the car when it is at the top of the circle?
Physics
1 answer:
11111nata11111 [884]2 years ago
7 0
Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m 
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N 
</span>
1) What is the magnitude of the normal force on the care when it is at the bottom of the circle. 

<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>

<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>

<span>Centripetal force = 7119.55 N </span>


<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>

<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>

<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>

√<span>(gr) </span>

√<span>(9.8 x 11.5) = 10.62 m/s</span>
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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

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\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

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